Boiling Point Calculator

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The formula for boiling point elevation is given below.

**$\Delta$ T = K**_{b}m

The boiling point of the solution is given by

**T**_{Solution} = T_{Pure solvent} + $\Delta$ T
**Step 1 :**

**Step 2 :**

Boiling point is defined as the temperature at which a liquid and the vapor above it are in equilibrium. At normal atmospheric pressure, the boiling point of a liquid will be the temperature at which the liquid is at equilibrium with the atmosphere above the liquid. Boiling points of solutions are higher that the boiling points of the pure solvents.

Another way to say this is that boiling point is the temperature at which molecules leave the liquid phase and are moved into gas phase.

The formula for boiling point elevation is given below.

The boiling point of the solution is given by

Read the problem and list out the values given.

Substitute the values in the formula to find out the unknown value.

**$\Delta$ T = K _{b}mT_{Solution} = T_{Pure solvent} + $\Delta$ T**

If 10.0g of NaCl is added to 6.00L of water and that the density of the solution is 1.00g/ml. Also assume that this very dilute solution behaes ideally. The value of Kb for water is 0.512

^{o}C/m**Step 1 :**Given data

K

_{b}= 0.512^{o}C/m

10.0g NaCl $\times \frac{1\ mol\ NaCl}{58.5\ g\ NaCl}$ = 0.171 mo NaCl

6.00 L water $\times \frac{1.00\ kg\ water}{L\ water}$ = 6.00 kg water

$\frac{0.171\ mol\ NaCl}{6.00\ kg\ water}$ = 0.0285 m NaCl**Step 2 :**Substitute the values in the formula.

**$\Delta$ T = K**_{b}m

$\Delta$ T = 0.512 $\times$ 0.0285m

$\Delta$ T = 0.03

^{o}C**Answer :**The answer if $\Delta$ T = 0.014

^{o}CA solution of 10.0g of sodium chloride is added to 100.0g of water in an attempt to elevate the boiling point. What is the boiling point of the solution?

**Step 1 :**Given data

Moles of NaCl = $\frac{10.0}{58.5}$ = 0.171 mol

Molarity = $\frac{0.171}{0.100}$ = 1.71 m**Step 2 :**For NaCl

Kb(water) = 0.52^{o}C/m

$\Delta$ Tb = Kbm

$\Delta$ Tb = (2)(0.52)(1.71)

$\Delta$ Tb = 1.78^{o}C

T

_{solution}= T_{b}(pure solvent) + $\Delta$T_{b}

T

_{solution}= 100^{o}C + 1.78

T

_{solution}= 101.78^{o}C**Answer :**The answer is T

_{solution}= 101.78^{o}C